Lipotso tse 3 mabapi le li-equation tsa tšimo ea motlakase
1. Bolo e tsamaisang motlakase e nang le radius ea 10 cm e na le tjhaja ea motlakase ea 500 μC. Lintlha tsa A, B, le C li lutse moleng le bohareng ba bolo bo hole ba 12 cm, 10 cm le 8 cm ka ho latellana ho tloha bohareng ba bolo. Bala matla a tšimo ea motlakase lintlheng tsa A, B, le C!
Tsejoa:
Radius ea bolo e tsamaisang (R) = 10 cm = 0.1 m
Tefiso ea motlakase (q) = 500 μC = 500 x 10 -6 C
r A = 12 cm = 0,12 m
r B = 10 cm = 0,1 m
r C = 8 cm = 0,08 m
Sephetho sa Coulomb (k) = 9 x 10 9
Ho batloa: Matla a tšimo ea motlakase ntlheng ea A (E A ), ntlheng ea B (E B ) le ntlheng ea C (E C )
tharollo:
a) Matla a tšimo ea motlakase ntlheng ea A
E A = kq / r A 2 = (9 x 10 9 )(500 x 10 -6 ) / (0,12) 2 = (4500 x 10 3 ) / 0,0144 = 312500 x 10 3 = 3,125 x 10 8 N/C
b) Matla a tšimo ea motlakase ntlheng ea B
E B = kq / r B 2 = (9 x 10 9 )(500 x 10 -6 ) / (0,1) 2 = (4500 x 10 3 ) / 0,01 = 450.000 x 10 3 = 4,5 x 10 8 N/C
c) Matla a tšimo ea motlakase ntlheng ea C
E C = 0 bakeng sa ho ba bolong.
2. Haeba tefiso ea teko ea 4 nC e beoa ntlheng, tefiso e ba le matla a 5 × 10 - 4 N. Boholo ba tšimo ea motlakase E ke bofe ntlheng eo?
Tse tsejoang:
Teko ea motlakase (q) = 4 nC = 4 x 10 -9 Coulomb
Matla a motlakase (F) = 5 × 10 -4 N
Ho batloa: Boholo ba tšimo ea motlakase (E)
tharollo:
E = F / q = (5 × 10 -4 ) / (4 x 10 -9 ) = 1,25 x 10 5 N/C
3. Litefiso tse peli q B = 12 μC le q C = 9 μC li behiloe likhutlong tsa khutlotharo e nepahetseng joalo ka ha ho bontšitsoe ho Setšoantšo. Fumana matla a tšimo ea motlakase a ikutloang ntlheng ea A!
Tse tsejoang:
Tefiso ntlheng ea B (qB) = 12 μC = 12 x 10 -6 C
Tefiso ntlheng ea C (qC) = 9 μC = 9 x 10 -6 C
Sephetho sa Coulomb (k) = 9 x 10 9
r AC = 4 cm = 0,04 m
r AB = 3 cm = 0,03 m
Oa Batla: matla a motlakase ntlheng ea A
tharollo:
E AC = kq / r 2 = (9 x 10 9 )(9 x 10 -6 ) / (0,04) 2 = 81 x 10 3 / 0,0016 = 5,0 x 10 7 N/C
E AB = kq / r 2 = (9 x 10 9 )(9 x 10 -6 ) / (0,03) 2 = 81 x 10 3 / 0,0009 = 9,0 x 10 7 N/C
